Question

срочно математика 10 класс

Answer 1

$\boxed {sin(90^\circ -a)=cosa}\; \; \; \to \; \; \; \; sin70^\circ =sin(90^\circ -20^\circ )=cos20^\circ \\\\\boxed {\; cos2a=2cos^2\frac{a}{2}-1\; }\; \; \; \; \boxed {\; cos2a=1-2sin^2\frac{a}{2}\; }\\\\\\\frac{1+cos20^\circ }{1-sin70^\circ }=\frac{1+cos20^\circ }{1-cos20^\circ }=\frac{2\, cos^210^\circ }{2\,. sin^210^\circ }=ctg^210^\circ$